Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{x + 1}{-5x + 25} \div \dfrac{x - 2}{x^2 - 7x + 10} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{x + 1}{-5x + 25} \times \dfrac{x^2 - 7x + 10}{x - 2} $ First factor the quadratic. $r = \dfrac{x + 1}{-5x + 25} \times \dfrac{(x - 5)(x - 2)}{x - 2} $ Then factor out any other terms. $r = \dfrac{x + 1}{-5(x - 5)} \times \dfrac{(x - 5)(x - 2)}{x - 2} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (x + 1) \times (x - 5)(x - 2) } { -5(x - 5) \times (x - 2) } $ $r = \dfrac{ (x + 1)(x - 5)(x - 2)}{ -5(x - 5)(x - 2)} $ Notice that $(x - 2)$ and $(x - 5)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ (x + 1)\cancel{(x - 5)}(x - 2)}{ -5\cancel{(x - 5)}(x - 2)} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $r = \dfrac{ (x + 1)\cancel{(x - 5)}\cancel{(x - 2)}}{ -5\cancel{(x - 5)}\cancel{(x - 2)}} $ We are dividing by $x - 2$ , so $x - 2 \neq 0$ Therefore, $x \neq 2$ $r = \dfrac{x + 1}{-5} $ $r = \dfrac{-(x + 1)}{5} ; \space x \neq 5 ; \space x \neq 2 $